Question

If the dissociation constants of two weak acids `HA_(1)` and `HA_(2)` are `K_(1)` and `K_(2)`, then the relative strengths of `HA_(1)` and `HA_(2)` are given by
A. `(K_(1))/(K_(2))`
B. `(K_(1)+sqrt(K_(2)))/(K_(2)+sqrt(K_(1)))`
C. `((K_(1))/(K_(2)))^(1//2)`
D. `((K_(1))/(K_(2)))^(3//2)`

Answer 1

Correct Answer - C
The dissociation of a weak acid is related to its `K_(a)` value as
`{:(,HA,hArr ,H^(+),+,A^(-)),("Initial conc. (M)",1.0,,0,,0),("Eqm. conc.",C(1-alpha),,Calpha,,Calpha):}`
or `K_(a)=(C alpha.C alpha)/(C(1-alpha))=(C alpha^(2))/((1-alpha))`
If `a lt lt 1` and `C=1` mol then
`K_(a)=alpha^(2)` or `alpha=(K_(a))^(1//2)`
Ratio of relative strengths of two acids depend on depends of dissociation i.e., `alpha`
`:. (alpha_(1))/(alpha_(2))=((K_(1))/(K_(2)))^(1//2)`